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Suppose that \(25 \%\) of people have a dog, \(29 \%\) of people have a cat, and \(12 \%\) of people own both. What is the probability that someone owns a dog or a cat?

Short Answer

Expert verified
The probability that someone owns either a dog or a cat is 0.42.

Step by step solution

01

Determine the individual probabilities

From the problem statement, the probability of owning a dog \( P(D) = 0.25 \). Similarly, the probability of owning a cat \( P(C) = 0.29 \). The probability of owning both a dog and a cat can be denoted as \( P(D \cap C) \) and is given as \( 0.12 \).
02

Apply the formula for the probability of the union of two events

The probability that someone owns a dog or a cat is given by the union of these two events. Therefore, we use the formula \( P(D \cup C) = P(D) + P(C) - P(D \cap C) \).
03

Substitute the values into the union formula

Substituting the given values into the formula results in: \( P(D \cup C) = 0.25 + 0.29 - 0.12 \)
04

Compute the result

Adding up the values gives \( P(D \cup C) = 0.42 \). This is the final probability that someone owns either a dog or a cat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probabilities of Individual Events
Understanding the probabilities of individual events is foundational to comprehending more complex probability concepts. In our exercise example, individual events include owning a dog and owning a cat. Here, we were given the probabilities of each event as 25% for dog ownership, denoted as \( P(D) = 0.25 \), and 29% for cat ownership, represented as \( P(C) = 0.29 \). These probabilities indicate the likelihood of each event occurring independently.

When we consider real-world scenarios, the likelihood of any event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In probabilistic terms, if we picture the entire population as a pie, the slices representing dog and cat owners are proportions of this pie, where the actual percentages are determined by surveys, studies, or other data collection methods.
Probability Formula Application
Applying probability formulas allows us to solve problems involving multiple events. The probability of the union of two events, which refers to the likelihood of at least one of the events occurring, can be calculated using the formula \( P(D \cup C) = P(D) + P(C) - P(D \cap C) \). This formula incorporates the concept of the addition rule in probability, accounting for the fact that some individuals may be counted in both individual event probabilities. Therefore, we subtract the probability of the intersection of the events to avoid double-counting those individuals.

The formula balances the total probability by ensuring that every individual is counted only once. As seen in the exercise, after inserting the given values, we can determine the combined probability of someone owning either a dog or a cat. The subtraction of the intersection probability, \( P(D \cap C) = 0.12 \), accounts for the overlap between dog and cat owners.
Intersection of Events
The intersection of events concept refers to the probability of two events occurring simultaneously. In our example, this is where the same individuals own both a dog and a cat. The intersection is denoted as \( P(D \cap C) \) and is given to be 12%. This figure is crucial as it represents the overlap between the two individual probabilities.

Understanding intersections is significant as it affects how we calculate combined probabilities. Neglecting to consider the intersection would result in an inflated probability, as people who have both a dog and a cat would be counted twice. Thus, by including the intersection in our calculations, as shown in the union formula, we arrive at a more accurate representation of the true probability.

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Most popular questions from this chapter

The soccer team's shirts have arrived in a big box, and people just start grabbing them, looking for the right size. The box contains 4 medium, 10 large, and 6 extra-large shirts. You want a medium for you and one for your sister. Find the probability of each event described. a. The first two you grab are the wrong sizes. b. The first medium shirt you find is the third one you check. c. The first four shirts you pick are all extra-large. d. At least one of the first four shirts you check is a medium.

You draw a card at random from a standard deck of 52 cards. Find each of the following conditional probabilities: a. The card is a heart, given that it is red. b. The card is red, given that it is a heart. c. The card is an ace, given that it is red. d. The card is a queen, given that it is a face card.

A survey found that \(73 \%\) of Americans have a home phone, \(83 \%\) have a cell phone and \(58 \%\) of people have both. a. If a person has a home phone, what's the probability that they have a cell phone also? b. Are having a home phone and a cell phone independent events? Explain. c. Are having a home phone and a cell phone mutually exclusive? Explain.

A nervous kicker usually makes \(70 \%\) of his first field goal attempts. If he makes his first attempt, his success rate rises to \(90 \%\). What is the probability that he makes his first two kicks?

Employment data at a large company reveal that \(72 \%\) of the workers are married, that \(44 \%\) are college graduates, and that half of the college grads are married. What's the probability that a randomly chosen worker a. is neither married nor a college graduate? b. is married but not a college graduate? c. is married or a college graduate?

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